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(A) The force $F$ is applied to the pulley. The tension in the string connected to the block $M$ is $F$. The string passes over the pulley,and one end

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$M g 	an \left(\frac{	heta}{2}ight)$

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(A) The force $F$ is applied to the pulley. The tension in the string connected to the block $M$ is $F$. The string passes over the pulley,and one end is fixed to the incline. The force $F$ acts at an angle $	heta$ with the incline. The component of $F$ along the incline is $F \cos 	heta$. The total force pulling the block up the incline is $T_{total} = F + F \cos 	heta$.For the block to move upward,this force must balance the component of gravity acting down the incline:$F + F \cos 	heta = M g \sin 	heta$$F(1 + \cos 	heta) = M g \sin 	heta$$F = \frac{M g \sin 	heta}{1 + \cos 	heta}$Using trigonometric identities $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$ and $1 + \cos 	heta = 2 \cos^2 \frac{	heta}{2}$:$F = \frac{M g (2 \sin \frac{	heta}{2} \cos \frac{	heta}{2})}{2 \cos^2 \frac{	heta}{2}}$$F = M g 	an \frac{	heta}{2}$

What is the minimum value of $F$ needed so that the block of mass $M$ begins to move upward on a frictionless inclined plane as shown?

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